#### Abstract

The concept of -distance was introduced in 2001; on the other hand, that of -function was introduced by Lin and Du. Strongly inspired by -function, we introduce a new concept, which is a very slight generalization of -distance and is more natural than -distance. So we could say that we redefine -distance in some sense.

#### 1. Introduction

Throughout this paper, we denote by , , and the sets of all positive integers, all rational numbers, and all real numbers, respectively.

In 2001, the concept of -distance was introduced in order to generalize results in [1–8] and others.

*Definition 1 (see [9]). *Let be a metric space. Then a function from into is called a -*distance* on if there exists a function from into and the following is satisfied:() for any .() and for any and , and is concave and continuous in its second variable.() and imply for any .() and imply .() and imply .

We note that the metric is one of -distances on . See [9–15] and references therein for many examples and theorems concerning -distance. For instance, using -distance, Suzuki [14] gave a simple proof of Zhong’s theorem [7].

In 2006, Lin and Du [16] introduced the following very interesting concept of -function, which is similar to that of -distance. However, both are independent.

*Definition 2 (Lin and Du [16]). *Let be a metric space. Then a function from into is called a *-function* if the following conditions hold: ()For any , .()If and in with and for some then .()For any sequence in with , and if there exists a sequence in such that , then .()For , and imply .

In this paper, strongly inspired by -function, we introduce a new concept, which is a slight generalization of -distance and which is more natural than -distance.

Before the author wrote the paper [9], he had considered more than twenty definitions. Definition 1 was the most natural definition between them. Therefore, he defined -distance as in Definition 1, though Definition 1 was not the weakest. However, strongly inspired by Definition 2, he finds a more natural definition of “-distance”; see Definition 7. Now he thinks that we have to change the definition of “-distance.”

#### 2. New Definition

In this section, we introduce a new definition of the notion of -distance. Before introducing it, we state something on -distance, which are strongly connected with the new definition.

*Definition 3 (see [9]). *Let be a -distance on a metric space . Then a sequence in is called *-Cauchy* if there exist a function from into satisfying (2)–(5) and a sequence in such that .

Lemma 4 (see [9]). *Let be a -distance on a metric space . If is a -Cauchy sequence, then is a Cauchy sequence in the usual sense.*

Lemma 5 (see [9]). *Let be a -distance on a metric space . If a sequence in satisfies for some , then is a -Cauchy sequence. Moreover, if a sequence in also satisfies , then . In particular for , and imply .*

Lemma 6 (see [9]). *Let be a -distance on a metric space . If a sequence in satisfies , then is a -Cauchy sequence. Moreover, if a sequence in satisfies , then is also a -Cauchy sequence and .*

Now we give the new definition.

*Definition 7. *Let be a metric space and let be a function from into . Then is called a -*distance* on if the following holds: () for any .()If and , then . Moreover, if converges to some , then for any .()If , then holds. Moreover, if converges to some , then for any .

The concept of -distance is slightly weaker than that of -distance. However, the author considers that the notions of both are the same. Indeed, we can prove -distance versions of all the existence theorems in [9, 11–14] using the same proof.

Proposition 8. *Let be a -distance on a metric space . Then is a -distance.*

*Proof. *() and () are the same. () follows from Lemma 6 and (). () follows from Lemmas 4 and 5 and (3).

*Remark 9. *A -distance need not be a -distance. However, the only example we have is quite complicated and we have chosen not to include it.

#### 3. Condition (CL)

We have introduced the concept of -Cauchy for -distance; see Definition 3. Instead of this, for -distance, we introduce Condition (CL), which is named after Cauchy and lower semicontinuity.

*Definition 10. *Let be a -distance on a metric space . Let be a net in . Then is said to satisfy* Condition (CL)* if the following holds:(CL1) is a Cauchy net in the usual sense.(CL2)Either of the following holds:(i) does not converge.(ii)If converges to , then holds for any .

The following lemma is important in this section.

Lemma 11. *Let be a metric space and let be a net in . Assume that is not Cauchy. Let be a sequence in and let be a mapping on . Then there exist a positive number and a sequence in such that for any .*

*Proof. *Since is not Cauchy, there exist a positive number and mappings and on such that , , and for any . Put and choose with and . Then either holds. So we can put or such that . Continuing this argument, we can obtain the desired result.

*Remark 12. * is not necessarily a subnet of .

We do not use the concept of net in Definition 7. However, we can prove something on net as follows.

Lemma 13. *Let be a -distance on a metric space . Let be a net in satisfying for some . Then the following holds: *(i)* satisfies Condition (CL).*(ii)*If a net in also satisfies , then holds.*

*Proof. *We choose a sequence in such that for . Arguing by contradiction, we assume is not Cauchy. Let be an arbitrary mapping from into itself. Then by Lemma 11, there exist a positive number and a sequence in satisfying for any . Since from (), we have which is a contradiction. Hence, is Cauchy. Next, we assume converges to . Fix and let with . Then we can choose a sequence in such that for any . Since converges to , we have from () Since is arbitrary, we obtain . Therefore, we have shown (i). Let us prove (ii). Define a directed set and an order ≤ in Define a net by Then since , is Cauchy by (i). This implies .

As a corollary of Lemma 13, we obtain the following sequential version. Compare Lemma 14 with Lemma 5.

Lemma 14. *Let be a -distance on a metric space . If a sequence in satisfies for some , then satisfies Condition (CL). Moreover, if a sequence in also satisfies , then holds. In particular, for , and imply .*

Lemma 15. *Let be a directed set and let be a mapping from into itself such that for any . Let be a net in a set . Then is a subnet of .*

*Proof. *Obvious.

Lemma 16. *Let be a -distance on a metric space . Let be a directed set such that for any there exists with . Let be a net in satisfying . Then the following holds: *(i)* satisfies Condition (CL).*(ii)*If a net in satisfies , then satisfies Condition (CL) and holds.*

*Proof. *We choose a sequence in such that for any . We consider the following two cases: (a)There exists such that for any .(b)For any , there exists with . In the first case, from the assumption, we can take with . Then since and for any , we have So Similarly, . By Lemma 14, we obtain for . Therefore, . Since converges to , thus, is Cauchy. Also, holds. Therefore, we have shown that satisfies Condition (CL) and . As in the above proof, we can prove that satisfies Condition (CL). In the second case, we can define a mapping from into itself satisfying for any . In fact, for , there exists with . We can choose satisfying and . If there exists such that and , then and hence , which is a contradiction. Similarly implies a contradiction. Therefore, we have defined . We will show that satisfies Condition (CL). Arguing by contradiction, we assume is not Cauchy. Then by Lemma 11, there exist a positive number and a sequence in satisfying From the definition of , we note for with . Since from (), we have So which is a contradiction. Hence, is Cauchy. Next, we assume converges to . Fix and let with . Then we can choose a sequence in such that Since , we have from () that Since is arbitrary, we obtain . Therefore, we have shown that satisfies Condition (CL). Let us prove . Arguing by contradiction, we assume . Fix with . Then we can choose a sequence in such that Since , we obtain from () thatwhich is a contradiction. Therefore, we have shown . We have proved (ii). We shall show that satisfies Condition (CL). Fix and let with . Then we can define a mapping from into itself such that By Lemma 15, we note that is a subnet of . We have So from (ii), satisfies Condition (CL) and holds. So, is Cauchy. We assume that converges to . Since also converges to , we have Since is arbitrary, we obtain . Therefore, we have shown that satisfies Condition (CL). We have proved (i).

As a corollary of Lemma 16, we obtain the following sequential version. Compare Lemma 17 with Lemma 6.

Lemma 17. *Let be a -distance on a metric space . If a sequence in satisfies , then satisfies Condition (CL). Moreover, if a sequence in satisfies , then satisfies Condition (CL) and holds.*

Lemma 18. *Let be a -distance on a metric space and let and be sequences in satisfying . Assume that, for any subsequence of the sequence in , there exist subsequences of and of satisfying . Then satisfies Condition (CL) and holds.*

*Proof. *We note that is Cauchy from Lemma 17. Let be an arbitrary subsequence of . Then from the assumption, there exist subsequences of and of satisfying . We note . From (), we have and hence Since is arbitrary, we obtain . is Cauchy and so is . Next, we assume that converges to , fix , and choose a subsequence of satisfying . Then from the assumption, there exist subsequences of and of satisfying . By (), we obtain Therefore, satisfies Condition (CL).

#### 4. Examples

In this section, in order to show that the concept of -distance is more natural than that of -distance, we give some examples. Compare them with Propositions and in [9] and Proposition in [11]. We note that we have not yet proved -distance versions of Propositions 19–21 below.

The following is connected with the result in Zhong [7]. See also Turinici [17].

Proposition 19. *Let be a -distance on a metric space . Let be a nonincreasing function from into with ; and let be a function from into with for . Then a function from into defined by for is also a -distance on .*

*Proof. *Using (), we have for any and hence holds. Define a function from into by for . From the assumption of , is well defined. If , then we have from the definition of . So, Hence, we note thatUsing (31), we next show . We assume and . Without loss of generality, we may assume and for any . Then for with , we have and hence This implies that is bounded. From the definition of , is also bounded and so is . We put Then we have by (31) By (), we obtain . Moreover, if converges to some , then for any . This yields for any . Therefore, holds. We can prove as in the proof of .

We call that a nonempty subset of is called *-bounded* if . The following is connected with the result in Ume [18].

Proposition 20. *Let be a -distance on a metric space . Let be a set-valued mapping on such that is a nonempty -bounded subset of for any . Assume that , , and imply . Then a function from into defined by for any is also a -distance.*

*Proof. *It is obvious that is -bounded for any . For nonempty -bounded subsets and of , we put Then . Indeed, for , we have Using , we can write From (), we have for nonempty -bounded subsets , , and of . So, for , we obtain This is . We next show . We assume and . We choose a subsequence of the sequence such that for any . Since , we can choose a sequence in such that for . We note Let be an arbitrary subsequence of and choose a subsequence of such that for any . Then we have and hence from Lemma 18. We also have and hence from Lemma 18. Thus, holds. If converges to some , then and also converge to . So from the assumption, holds. Also, holds for any from Lemma 18. We obtain We have shown . Let us prove . We assume . Fix . Then since satisfies Condition (CL) by Lemma 14. We assume that converges to . We choose a sequence in such that for . Since we have by Lemma 14. Hence, converges to . From the assumption, holds. As in the proof of (), we can prove for any .

We finally prove the following. We note that our proof is very easy while the proof of Proposition in [11] is difficult.

Proposition 21. *Let be a metric space. Let be a set of -distances on . If a function defined by is a real-valued function, then is a -distance on .*

*Proof. *We have for any . We next show . We assume and . Then since for any , and hold. Thus, from , holds. If converges to some , then holds for any and by using again. Using it, we have for any . We have shown . We can similarly prove .

#### 5. Kirk and Saliga’s Fixed Point Theorem

Kirk and Saliga generalized Caristi’s fixed point theorem [2, 3]; see Theorem in [19]. The author thinks that the proof in [19] is splendid. In this section, we generalize Kirk and Saliga’s fixed point theorem.

Let be an ordinal number. We denote by and the successor and the predecessor of , respectively. We recall that is* isolated* if exists. is* limit* if and does not exist. For a set , we denote by cardinal number of .

Theorem 22. *Let be a complete metric space with a -distance . Let be a mapping on . Let be a function from into bounded from below such that whenever a net satisfies the following: *(a)* converges to .*(b)*For any , there exists satisfying .*(c)*.*(d)* for with .**Assume for any , and there exists such that for any . Fix . Let be the first uncountable ordinal number and put . Define a net by Then is well defined and there exists satisfying and .*

*Proof. *Define a function form into by where is the identity mapping on . It is obvious that holds for any . We shall show by transfinite induction that (a) is well defined,(b) for ,for . It is obvious that is true. Fix with , and assume that is true for . In the case where is isolated, it is obvious that is well defined. We have For , we have Hence, is true. In the other case, where is limit, since is a nonincreasing and is bounded from below, converges and hence it is Cauchy. We have By Lemma 16, satisfies Condition (CL). Since is complete, converges. Hence, is well defined. For , we have Hence, is true. Therefore, by transfinite induction, is true for any . Arguing by contradiction, we assume the following: (a) for any .Then there exists satisfying . So we have which is a contradiction. Therefore, there exists satisfying . We have Hence, and hold. By Lemma 14, holds. Therefore, holds.

#### Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.

#### Acknowledgments

The author is supported in part by Grants-in-Aid for Scientific Research from the Japanese Ministry of Education, Culture, Sports, Science and Technology.